Legendre's Condition

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Theorem

Let $y =\map y x$ be a real function, such that:

$\map y a = A,\quad \map y b = B$

Let $J\sqbrk y$ be a functional, such that:

$\displaystyle J\sqbrk y = \int_a^b \map F {x,y,y'}\rd x$

where

$F \in C^2\closedint a b$

with respect to all its variables, and $C$ stands for differentiability class.


Then a necessary condition for $J\sqbrk y$ to have a minimum at $y = \hat y$ is

$F_{y'y'} \big \vert_{y \mathop = \hat y} \ge 0 \quad \forall x \in \closedint a b$.


Proof

Lemma 1

Let $y = \map y x$ be a real function, such that:

$\map y a =A,\quad\map y b = B$

Let $J \sqbrk y$ be a functional, such that:

$\displaystyle J \sqbrk y=\int_a^b \map F {x,y,y'} \rd x$

where

$F\in C^2\closedint a b$

with respect to all its variables.


Then

$\displaystyle\delta^2 J\sqbrk{y;h}=\int_a^b\paren{\map P {x,\map y x }h'^2+\map Q {x,\map y x}h^2}\rd x$

where

$\displaystyle P\paren{x,\map y x}=\frac 1 2 F_{y'y'},\quad \map Q {x,\map y x}=\frac 1 2 \paren {F_{yy}-\frac \d {\d x}F_{yy'} }$


Lemma 2

Let $h$ be a real function such that:

$h\in C^1 \openint a b,\quad\map h a=0,\quad \map h b=0$

Let

$\displaystyle \delta^2 J \sqbrk{y;h} = \int_a^b \paren{\map P {x,\map y x}h'^2 + \map Q {x,\map y x} h^2}\rd x$

where $P\in C^0\closedint a b$.


Then a necessary condition for

$\delta^2 J\sqbrk{y;h}\ge 0$

is

$\map P {x,\map y x}\ge 0\quad\forall x \in \closedint a b$

$\blacksquare$


Source of Name

This entry was named for Adrien-Marie Legendre.


Sources